Question

what multiplies to -1728 and adds to -48?

Answer 1

Since the product is negative, one number will be negative and the other positive.

Lets do some algebra.
Let x = one number and let y = the other now express the problem in terms of x and y
xy= -1728
x+y= -48 we have two equations with two unknowns, a good situation.

Answer 2

Now from above we can express x in terms of y or x = -48 - y (using the second equation)
Now substitute that for x in the first equation, getting (-48 - y)y=1728. Expane the equation getting-y^2-48y = -1728. Now express as a quadratic
-y^2 - 48y + 1728 = 0 Multiply all terms by -1 getting:
y^2 +48y - 1728 =0 Now solve the quadratic the same as you have been solving quadratics.

Answer 3

i understand a little bit of that haha :/

Answer 4

at the risk of being a jerk, unless you are going to solve this using the quadratic equation, it is still going to come down to the original question

Answer 5

you might want to consider factors of \(1728\) and see which two have a difference of \(48\)
or if you simply want to cheat, solve
\[x^2-48-1728=0\] for \(x\) using the quadratic formula, and a calculator

Answer 6

oops i mean
\[x^2-48x-1728=0\]

Answer 7

great idea thanks :) my brain is dead at this point

Answer 8

*error on my previous post. left off minus sign, should read: (-48 - y)y=-1728*

Answer 9

Please forgive, it was a typo.!

Answer 10

Remainder of post was O.K.

Answer 11

@radar the problem is that when you set up the quadratic equation you will come back to the same question again if you try to factor. the advantage of your method however is that you can always solve a quadratic equation by using the quadratic formula
this is particularly ugly because \(1728=2^6\times 3^3\) and so it has many factors

Answer 12

I used the quadratic and came up with -72 and 24 for a solution. However, I was solving for y my quadratic was:\[y ^{2} + 48y-1728=0\] Which can be factored (y+72)(y-24)

Answer 13

\[y_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-1728)}}{2(1)}\]